Question

Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd? (Enter your probability as a fraction.)

Answer 1

6 of the 10 integers between 1 and 10 inclusive are more than 4 (that is, 5,6,7,8,9,10).
The probability is the number of favorable outcomes divided by the number of possible outcomes: \(P(>4)=\# \)of favorable outcomes \(\#\) of possible outcomes=\(\frac{6}{10}\)
5 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=# of favorable outcomes/# of possible outcomes=\(\frac{5}{10}\)
3 of the 10 integers between 1 and 10 inclusive are move than 4 and odd (that is, 5,7,9). \(P(>4\) and odd\()=\#\) of favorable outcomes/# of possible outcomes=\(\frac{3}{10}\)
Use the General addition rule for any two events: \(P(AUB)=P(A)+P(B)-P(A⋂B)\)
P(>4 or odd) \(=P(>4)+P(odd)-P(>4 \text{and odd}) =\frac{6}{10}+\frac{5}{10}-\frac{3}{10} =\frac{6+5-3}{10} =\frac{8}{10} =\frac{4}{5}\)